Expected value of $X^{-1}$, $X$ being a noncentral $\chi^2$. Cannot understand a step of a equation in a paper I am reading the following paper:
Mudholkar GS, Chaubey YP, Ching-Chuong L (1976). Approximations for the doubly noncentral-$F$ distribution. Communications in Statistics - Theory and Methods, 5(1):49–63. doi:10.1080/03610927608827331
In this paper (section 2, page 51), $X$ and $Y$ are both defined to be noncentral $\chi^2$ random variables, with respective degrees of freedom $\nu_1$ and $\nu_2$ and noncentrality parameters $\lambda_1$ and $\lambda_2$. $X$ and $Y$ are also assumed to be independent.
The authors try to obtain an approximation to the raw moments of the ratio $X/Y$ (which is known to follow a doubly noncentral $F$ distribution, given the assumptions stated above).
The $r$-th raw moment $\mu'_r$ of $X/Y$ is defined as $E\left[\left(\frac{X}{Y}\right)^r\right]$. Using the independence of $X$ and $Y$, the first equality in equation (2.1) of the paper says (as far as I understand):
$$
\mu'_r = E\left[X^r\right] E\left[Y^{-r}\right] \, \mathrm{,}
$$
which makes sense to me.
My first doubt is a notation issue. The previous equation is actually typed in the paper as follows:

$$\mathtt{\mu'_r=EX^r \cdot EY^{-r}}$$

(using a typewriter font, as all the paper).
Later in the same page, the authors include this notation explanation:

$\mathtt{\mu_Y=EY}$

So I guess that $\mathtt{EY^{-r}}$ stands for $E\left[Y^{-r}\right]$, doesn't it?
The previous equation is further developed like this in the paper:

$$\mathtt{\mu'_r=EX^r \cdot EY^{-r} = EX^r \cdot E \left[ 1 + \frac{Y-EY}{EY} \right]^{-r} } \, \mathrm{.}$$

And, maybe it is very easy, but it is here where I get totally lost. I do not understand this last step. Could you please give me some light?

NOTE: According to this thread: Is it okay to ask question about a specific paper / model?, it is OK to post questions on specific papers like the present one.

UPDATE
Just (hoping) to make it clearer, the whole equation (2.1) in the article is like follows (using their notation):

$$
\begin{align*}
\mathtt{\mu'_r} &\mathtt{= EX^r \cdot EY^{-r}} \\
                &\mathtt{= EX^r \cdot E \left[ 1 + \frac{Y-EY}{EY} \right]^{-r}}\\
                &\mathtt{= EX^r \cdot \left[ 1 + \binom{-r}{2}\frac{\mu_{2,Y}}{\mu_Y^2} + \binom{-r}{3}\frac{\mu_{3,Y}}{\mu_Y^3} + \binom{-r}{4}\frac{\mu_{4,Y}}{\mu_Y^4} + \cdots \right]}
\end{align*}
$$

where $\binom{-r}{k}$ stands for  $\,\prod_{j=1}^{k}{\frac{-r-j+1}{j}}\,$, $\,\mu_Y = E[Y]\,$ and  $\,\mu_{r,Y} = E\left[\left( Y - \mu_Y \right)^r\right]\,$ (central moments).
 A: Yes, $EY^{−r}$ stands for $E[Y^{−r}]$. (I dislike not making it explicit because it leaves too many opportunities for misunderstandings and errors.)
With respect to the later part, consider:
$Y=\frac{Y}{E(Y)}\cdot E(Y)= E(Y)\cdot [\frac{Y}{E(Y)}-1+1]= E(Y)\cdot [\frac{Y-E(Y)}{E(Y)}+1]$
Therefore

$EY^{-r} = [E(Y)]^{-r}\cdot E \left[ \left( 1 + \frac{Y-EY}{EY} \right)^{-r}\right]  \, .$

So now compare with the paper. The equation in the paper is 

$\mu'_r=EX^r \cdot EY^{-r} = EX^r \cdot E \left[ 1 + \frac{Y-EY}{EY} \right]^{-r}\,.$ 

Dividing through by $EX^r$ we have 

$\mu'_r/EX^r =  EY^{-r} = E \left[ 1 + \frac{Y-EY}{EY} \right]^{-r}$. 

So we see from that last equality that they're asserting

$EY^{-r} = E \left[ 1 + \frac{Y-EY}{EY} \right]^{-r}\quad{^\ddagger}$

and so we can see $[E(Y)]^{-r}$ seems to have disappeared -- there's a term missing in the paper. 
$^\ddagger$ -- keeping in mind that when they write  $E\text{<term>}^{-r}$ it seems they intend $E(\text{<term>}^{-r})$, so this means $E\left( \left[ 1 + \frac{Y-EY}{EY} \right]^{-r}\right)$ 
